On 10/06/2013 05:52 AM, David Bernier wrote:> On 10/02/2013 10:29 PM, Mike Terry wrote:>> "quasi" <quasi@null.set> wrote in message>> news:v9vo49lvr7j4ns7i82s869ou749lgdmrup@4ax.com...>>> Pubkeybreaker wrote:>>>> Susam Pal wrote:>>>>>>>>>> How can you construct a plane where every point is coloured>>>>> either black or white such that two points of the same colour>>>>> are never a unit distance apart?[snip]>>>> It does not seem possible. Take any point. Suppose it is>>>> white. It is the center of a circle (of radius 1). All points>>>> on that circle must then be black. But given any point on>>>> that circle there is another point on the circle at distance 1.>>>> Both points are the same color.>>>>>>>> Unless, of course, I am being completely stupid.>>>>>> No, your proof is fine.>>>>>> quasi>>>> The question was asked here a few years ago whether the plane can be>> coloured along the lines above but using three colours. The answer is no,>> and the proof is similar but less obvious...> [...]> > There is a web-page called "Chromatic Number of the Plane" by> Alexander Bogomolny that briefly discusses the question of> the minimum number of colours needed.> > The relevant definition, copied from there, is:> ``The smallest number of colors needed in a coloring of the plane to> ensure that no monochromatic pair is at the unit distance apart is> called the chromatic number Chi of the plane."> > Ref.:> < http://www.cut-the-knot.org/proofs/ChromaticNumber.shtml > .> > Two or three colours won't do, from which we see that Chi >= 4.> A 7-colouring of a regular-hexagon tiling of the plane shows> that seven colours will do, from which we see that Chi <= 7.

[snip]

> The Chi question, according to the Wikipedia article above, is> also known as the Hadwiger?Nelson problem :> > < http://en.wikipedia.org/wiki/Hadwiger%E2%80%93Nelson_problem >> > The "Moser spindle" unit-distance graph shows that Chi >=4 :> < http://en.wikipedia.org/wiki/Moser_spindle >> > There are only 7 vertices, so it's easy to describe:> > Let A be a point in the plane at (0, 0).> Let B be the point (1,0).> Let C be the point in Quadrant I which is at unit distance> from both A and B. So C has coordinates: (cos pi/3, sin pi/3).> > A is one of the two points which are at unit distance from> both B and C. Let us call the second point D.> > Let E be an as yet undetermined point on the circle of> radius 1 centered at A.> > Let F be determined, relative to A and E, by requiring> that the unit-segment AF is obtained by rotating the> unit-segment AE through pi/3 [60 degrees] counter-clockwise.> > Then, E and F shall be one unit distance apart.> > A is one of the two points at unit distance from both E and> F. Let G be the second point at unit distance from both E and> F.> > We have the unit-sided equilateral triangles: ABC, BCD,> AEF, EFG.> > Like ("suggested") in the figure in the top right box at:> http://en.wikipedia.org/wiki/Moser_spindle ,> > with A being the highest point in the figure,> points D and G are the furthest from A,> and also the lowest in the drawing.> > The points called B, C, E, F are all at unit distance> from A .> > Then we require finally that D and G be a unit distance apart.> Then the line segments AD and AG have the same length.> We can fix G down to unique by asking that there be a> counter-clockwise rotation through A of less than 180 degrees> that maps segment AD to segment AG.> > With a three-colouring, considering in turn> {A, B, C, D} and {A, E, F, G},> one is forced to colour D the colour of A, and also> forced to colour G the colour of A. So D and G, although> unit distance apart, share the same colour (X).> > So the Moser spindle needs at least 4 colors.> > The Moser spindle is related by Wikipidia to be a Laman graph,> which have a type of rigidity property:> < http://en.wikipedia.org/wiki/Laman_graph > .> > After some thinking, I realized that the Moser spindle, as a rigid> figure in the plane consisting of 4 equilateral triangles of> unit length and 7 distinct vertices, is constructible by> ruler and compass ...> > Also, I tried that out experimentally.> > The wikipedia article on ruler and compass,> < http://en.wikipedia.org/wiki/Compass-and-straightedge_construction >> > says that:> ``It is possible (according to the Mohr?Mascheroni theorem) to construct> anything with just a compass if it can be constructed with a ruler and> compass, provided that the given data and the data to be found consist> of discrete points (not lines or circles)."> > In the Moser spindle case, if we suppose A = (0, 0) and> B = (1, 0) as the starting point,> then it's easy to construct C by compass.> > Also, constructing D by compass is easy. Then G seems> easy also, which leaves E and F.> > So, I thought of writing a computer program to devise pseudo-random> compass-only constrcutions of finite number of points, with an aim> of getting "lots" of pairs of points at unit distance from each other.[snip]

(aside: Erdos wrote that he was almost sure that Chi > 4, for the plane).

I thought of a decision problem, related to "random graphs"with many pairs of points at unit distance from each other.

Suppose I have a 50x50 matrix T with entries in {0, 1}.It's a given that T is symmetric, meaning T = T^t, andthat the main diagonal of T is all zeros.